Floor sum of numbers

[man111]

The sum of series [tex]\displaystyle 3\lfloor \frac{1}{3!}+4\lfloor \frac{1}{4!}+.......+100\lfloor \frac{1}{100!}\rfloor\rfloor\rfloor[/tex], where [tex]\lfloor x \rfloor[/tex] floor function of [tex]x[/tex]

[nevrodermit]

Note that
[tex]100\lfloor \frac{1}{100!} \rfloor = 0 \Rightarrow 99 \lfloor \frac{1}{99!}+100\lfloor \frac{1}{100!} \rfloor \rfloor = 99\lfloor \frac{1}{99!} \rfloor = 0[/tex]
Continuing this way we get
[tex]3 \lfloor \frac{1}{3!}+0 \rfloor = 3\lfloor \frac{1}{3!} \rfloor = 0[/tex]